\begin{align} & \left( {{\text{A + B}}} \right){\text{'s}}\, {\text{20}}\, {\text{day's}}\, {\text{work}} \cr & = \left( {\frac{1}{{30}} \times 20} \right) = \frac{2}{3} \cr & {\text{Remaining}}\, {\text{work}} \cr & = \left( {1 - \frac{2}{3}} \right) = \frac{1}{3} \cr & {\text{Now}}, \frac{1}{3}\, {\text{work}}\, {\text{is}}\, {\text{done}}\, {\text{by}}\, {\text{A}}\, {\text{in}}\, {\text{20}}\, {\text{days}} \cr & \therefore {\text{The}}\, {\text{whole}}\, {\text{work}}\, {\text{will}}\, {\text{be}}\, {\text{done}}\, {\text{by}}\, {\text{A}}\, {\text{in}} \cr & \left( {20 \times 3} \right) = 60\, days \cr\end{align}

M₁ *D₁ = M₂ *D₂

mr = (m +n) *D₂

D₂ = ^mr/_{(m +n)} .

\begin{align} & \left( {{\text{A + B}}} \right){\text{'s}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = \frac{1}{{10}} \cr & {\text{C's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = \frac{1}{{50}} \cr & \left( {{\text{A + B + C}}} \right){\text{'s}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} \cr & = \left( {\frac{1}{{10}} + \frac{1}{{50}}} \right) = \frac{6}{{50}} = \frac{3}{{25}}.....\left( {\text{i}} \right) \cr & {\text{A's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} \cr & = \left( {{\text{B + C}}} \right){\text{'s}}\, {\text{1}}\, {\text{day's}}\, {\text{work}}\, .....\left( {{\text{ii}}} \right) \cr & {\text{From}}\, \left( {\text{i}} \right)\, {\text{and}}\, \left( {{\text{ii}}} \right){\text{, we}}\, {\text{get}}:2 \times \left( {{\text{A's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}}} \right) \cr & = \frac{3}{{25}} \cr & \Rightarrow {\text{A's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = \frac{3}{{50}} \cr & \therefore {\text{B's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}}\left( {\frac{1}{{10}} - \frac{3}{{50}}} \right) \cr & = \frac{2}{{50}} = \frac{1}{{25}} \cr & {\text{So, }}\, {\text{B}}\, \, {\text{alone}}\, {\text{could}}\, {\text{do}}\, {\text{the}}\, {\text{work}}\, {\text{in}}\, {\text{25}}\, {\text{days}} \cr\end{align}

Difference in times required by the first man (A) and second man (B) = 3 hours. Also, if t_a and t_b are the respective times, then

t_b - t_a = 3 . . . . . . . . . ..(1)

Also, B alone be take = (t_a + 3) h

According to the question,

2t_b - t_a = 8

2* (t_a + 3) - t_a = 8 [Using equation (1)]

t_a = 2 hours.

Now B and woman together take 2 hours and A also take 2 hours, so time required will be half when all 3 work together. So in 1 hour work would be completed.

Working efficiency of both typist together,

= ¹⁰⁰/₆ = 16.66% per minute

Now, let work efficiency of first typist be x and then second typist will be (16.66 - x)

First typist typed alone for 4 minutes and second typed alone for 6 minutes and they left with ¹/₅ (i.e 20%) of job, means they have completed 80% job.

Now,

First Typist typed in 4 minute + Second typed in 6 minutes = 80%

4 *x + 6 *(16.66 - x) = 80%

4x + 100% - 6x = 80%

x = 10%

First Typist typed 10% per minutes. Then second typed (16.66 - 10) = 6.66% per minute

Then, Second typist complete the whole job in ¹⁰⁰/_6.66 = 15.01 = 15 minutes.